Is the set of finite rank operators with rank at most n closed w.r.t the weak operator topology?

Question

I know the set of finite rank operators with rank less than n is closed in strong operator topology. Can we say it is also closed in weak operator topology?

Answer 1

EDIT: There is a very short proof in case you have the result for one of the mentioned operator topologies available: SOT (Strong Operator Topology) and WOT (Weak Operator Topology) generate the same topological dual space and thus convex sets are SOT-closed iff they are WOT-closed. Note that linear subspaces of vector spaces are convex and thus the desired result follows immediately since the set of all operators of rank at most $k$ is a linear subspace.

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Now to the original proof I provided:

For all $k \in \mathbb{N}$, let $ \mathfrak F_k(X,Y) = \mathfrak F_k$ denote the set of finite rank operators with rank at most $k$.

Showing that $\mathfrak F_k$ is closed w.r.t the strong operator topology (SOT) is similar to showing that $\mathfrak F_k$ is closed w.r.t the weak operator topology (WOT) since on normed spaces of finite dimension norm convergence and weak convergence are equivalent.

We start by proving the following

Lemma. Let $(w_1^i)_{i \in I}, \dots, (w_k^i)_{i \in I} \subseteq Y$, $k \in \mathbb{N}$, be nets converging weakly to the linearly independent vectors $w_1,\dots,w_k$. Then there exists $i_0 \in I$ such that for all $i \geq i_0$ the vectors $w_1^i, \dots, w_k^i$ are linearly independent.

Proof. Consider the linear subspace $V:= \langle w_1, \dots, w_k\rangle$ and the corresponding dual basis $w_1^*, \dots, w_k^*$. Since, $V$ is finite dimensional $w_1^*, \dots, w_k^*$ can be extended continuously to functionals on $Y$ (Hahn-Banach) which we will again denote by $w_1^*,\dots,w_k^*$.

We can now explicitly write down the continuous projection $p_V$ onto $V$: $$ p_V(x) = w_1^*(x) w_1 + \dots + w_k^*(x) w_k. $$ As $w_1^i,\dots,w_k^i$ converge weakly, their images under $p_V$ will converge strongly. You can check out e.g. this post to see that the vectors $p_V(w_1^i),\dots, p_V(w_k^i)$ will eventually become linearly independent and so will their preimages $w_1^i,\dots,w_k^i$.$\quad\square$

Now, let $(f_i)_{i \in I} \subseteq \mathfrak{F}_k$ be a SOT-convergent net. Suppose $f := \operatorname{WOT-}\lim_i f_i \not \in \mathfrak{F}_k$. Then, there exist linearly independent vectors $\eta_1, \dots, \eta_{k+1} \in \operatorname{range} f$ with corresponding linearly independent preimages $\xi_1, \dots, \xi_{k+1}$, i.e. $$ f(\xi_{1}) = \eta_1,\dots, f(\xi_{k+1}) = \eta_{k+1}. $$ Since $(f_i(\xi_j))_{i \in I}, j = 1,\dots,{k+1},$ converge weakly, by the above lemma there exists $i_0 \in I$ such that $f_{i_0}(\xi_1),\dots,f_{i_0}(\xi_{k+1})$ are linearly independent. This contradicts the assumption that all $f_i$ were of at most rank $k$. Thus, we have proved that $$ \mathfrak F_k = \overline{\mathfrak F_k}^{\operatorname{WOT}} \left(= \overline{\mathfrak F_k}^{\operatorname{SOT}} \right) $$ for all $k \in \mathbb{N}$.

Answer 2

Lemma: Let $X$ be a normed space. Suppose $\{\xi_1, \dots, \xi_n\}$ is a linearly independent set, and for each $1 \leq i \leq n$, choose a functional $f_i \in X^*$ such that $$ f_i(\xi_j) = \delta_{i,j}. $$ Then there exists $\varepsilon > 0$ such that for any choice of $$ \eta_j \in B_j \equiv \left\{ \eta \in X : \bigl|f_i(\eta - \xi_j)\bigr| < \varepsilon,\ \text{for all }1 \leq i \leq n\right\} $$ (note that $B_j$ is a weak neighborhood of $\xi_j$), the set $\{\eta_1, \dots, \eta_n\}$ remains linearly independent.

Proof. Assume the conclusion is false.

Claim. For each $\varepsilon > 0$, there exist scalars $\lambda_1^\varepsilon, \dots, \lambda_n^\varepsilon \in \mathbb{C}$ such that $$ \sum_{j=1}^n \bigl|\lambda_j^\varepsilon\bigr| = 1 \quad\text{and}\quad \Bigl|\,f_i\bigl(\sum_{j=1}^n \lambda_j^\varepsilon \xi_j\bigr)\Bigr| < \varepsilon, \quad \text{for all }1 \leq i \leq n. $$

Indeed, since the conclusion fails, for each $\varepsilon>0$ there are points $$ \eta_j^\varepsilon \in B_j \equiv \left\{ \eta \in X : \bigl|f_i(\eta - \xi_j)\bigr| < \varepsilon,\ \forall\,1\le i\le n\right\} $$ making $\{\eta_1^\varepsilon,\dots,\eta_n^\varepsilon\}$ linearly dependent. Hence there exist scalars $\alpha_1^\varepsilon,\dots,\alpha_n^\varepsilon$ such that $$ \sum_{j=1}^n \alpha_j^\varepsilon\,\eta_j^\varepsilon = 0 \text{ and }\sum_{j=1}^n |\alpha_j^\varepsilon|>0. $$ Then for each $1\le i\le n$, $$ \Bigl|f_i\bigl(\sum_{j=1}^n \alpha_j^\varepsilon\xi_j\bigr)\Bigr| = \Bigl|f_i\bigl(\sum_{j=1}^n \alpha_j^\varepsilon(\xi_j-\eta_j^\varepsilon)\bigr)\Bigr| < \varepsilon \sum_{j=1}^n \bigl|\alpha_j^\varepsilon\bigr|. $$ Setting $$ \lambda_k^\varepsilon = \frac{\alpha_k^\varepsilon}{\sum_{j=1}^n \bigl|\alpha_j^\varepsilon\bigr|}, \qquad 1\le k\le n, $$ gives the Claim.

Now let $$ \mathbb{T} = \bigl\{(\lambda_1,\dots,\lambda_n)\in\mathbb{C}^n : \sum_{j=1}^n|\lambda_j|=1\bigr\}, $$ a compact subset of $\mathbb{C}^n$, and define $$ \varphi(\lambda_1,\dots,\lambda_n) = \max_{1\le i\le n} \Bigl|\,f_i\bigl(\sum_{j=1}^n\lambda_j\xi_j\bigr)\Bigr|. $$ By the Claim, $$ \inf_{(\lambda_1,\dots,\lambda_n)\in\mathbb{T}}\varphi(\lambda_1,\dots,\lambda_n)=0. $$ By the continuity of $\varphi$ and the compactness of $\mathbb{T}$ yields $(\lambda_1^0,\dots,\lambda_n^0)\in\mathbb{T}$ with $\varphi(\lambda_1^0,\dots,\lambda_n^0)=0$, that is $$ \lambda_i=f_i\Bigl(\sum_{j=1}^n\lambda_j^0\xi_j\Bigr)=0 \text{ for }1\leq i\leq n,\quad \sum_{j=1}^n|\lambda_j^0|=1, $$ a contradiction.

Theorem: Let $X$ be a normed space and $n\in\mathbb{N}$. Then the set of operators in $B(X)$ with rank at most n is WOT closed.

Proof. Suppose $T\in B(X)$ has $\operatorname{rank}(T)>n$. Then there exist $$ \xi_1,\dots,\xi_{n+1}\in X \quad\text{with}\quad \{T\xi_1,\dots,T\xi_{n+1}\}\text{ linearly independent}. $$ By the Lemma, pick $f_i\in X^*$ so that $f_i(T\xi_j)=\delta_{i,j}$ and let $\varepsilon>0$ ensure that any $$ \eta_j\in B_j\equiv\{\eta:|f_i(\eta-T\xi_j)|<\varepsilon,\ 1\leq i\le n+1\} $$ yields $\{\eta_1,\dots,\eta_{n+1}\}$ linearly independent. Then $$ N = \bigl\{S\in B(X):S\xi_j\in B_j,\ 1\leq j\le n+1\bigr\} = \bigl\{S\in B(X):|f_i((S-T)\xi_j)|<\varepsilon,\ 1\leq i,j\leq n+1\bigr\} $$ is a WOT‐neighborhood of $T$, and every $S\in N$ satisfies $\operatorname{rank}(S)>n$, since $\{S\xi_1,\dots,S\xi_{n+1}\}$ is linearly independent. This shows that the set of operators with rank strictly greater than $n$ is WOT-open. Thus, the set of operators with rank at most $n$ is WOT-closed.